What are wrong limits?

Choose α and β so the curve sweeps the region exactly once; use symmetry to integrate over half and double, but only if the region is genuinely symmetric.

Find the area enclosed by r = 2 for 0 θ 2π. [2 marks]

Convert the point with polar coordinates (2, (π)/(3)) to Cartesian form. [2 marks]

Convert r = 3cosθ to Cartesian form. [3 marks]

Pearson Edexcel Edexcel 2019-style practice: The curve C has polar equation r = a(1 + cosθ) for 0 θ 0. Find the exact area enclosed by C.

Apply the polar area formula and use the double-angle identity to integrate cos^2θ. A = (1)/(2)_0^2π r^2\,dθ = (1)/(2)_0^2π a^2(1 + cosθ)^2\,dθ (M1). Expand: (1 + cosθ)^2 = 1 + 2cosθ + cos^2θ = 1 + 2cosθ + (1 + cos 2θ)/(2) (M1 A1). So A = (a^2)/(2)_0^2π((3)/(2) + 2cosθ + (1)/(2)cos 2θ)dθ. Over 0 to 2π the cosθ and cos 2θ terms integrate to zero (M1 A1), leaving A = (a^2)/(2)·(3)/(2)· 2π = (3π a^2)/(2) (A1).

Pearson Edexcel Edexcel 2021-style practice: The curve has polar equation r = 1 + 2cosθ. Find the polar coordinates of the points where the tangent to the curve is perpendicular to the initial line.

Tangents perpendicular to the initial line occur where (d)/(dθ)(rcosθ) = 0. rcosθ = (1 + 2cosθ)cosθ = cosθ + 2cos^2θ (M1). Differentiate: (d)/(dθ)(rcosθ) = -sinθ - 4cosθsinθ = -sinθ(1 + 4cosθ) (M1 A1). Set to zero: sinθ = 0 or cosθ = -(1)/(4) (A1). Taking cosθ = -(1)/(4) gives θ = (-14) ≈ 1.823 and its reflection, with r = 1 + 2(-14) = 12 (A1).

Pearson Edexcel Edexcel 2023-style practice: Convert the polar equation r = 4sinθ to Cartesian form and describe the curve.

Multiply by r to introduce r^2 and rsinθ, then substitute. r = 4sinθ r^2 = 4rsinθ (M1). Using r^2 = x^2 + y^2 and rsinθ = y: x^2 + y^2 = 4y (A1). Completing the square: x^2 + (y - 2)^2 = 4, a circle of radius 2 centred at (0, 2) (A1).

EnglandFurther MathsSyllabus dot point

How do polar coordinates describe curves, and how do you find areas they enclose?

Polar coordinates and curves, conversion to and from Cartesian form, sketching cardioids and spirals, tangents parallel and perpendicular to the initial line, and areas enclosed by polar curves.

A focused answer to the Edexcel A-Level Further Mathematics polar coordinates content, covering polar coordinates and curves, conversion between polar and Cartesian form, sketching cardioids and spirals, finding tangents parallel and perpendicular to the initial line, and computing areas enclosed by polar curves.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Polar coordinates and conversion
Sketching polar curves
Tangents and area
Examples in context
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What this dot point is asking

Edexcel wants you to plot points and curves in polar form $(r, \theta)$ , convert between polar and Cartesian equations, sketch standard polar curves such as cardioids and spirals, find tangents parallel and perpendicular to the initial line, and use the polar area formula. Polar work is a self-contained Core Pure topic that rewards careful sketching and clean trigonometric integration.

Polar coordinates and conversion

In polar form a point is given by its distance $r$ from the pole (origin) and the angle $\theta$ measured anticlockwise from the initial line (the positive $x$ -axis). The conversion to Cartesian uses basic right-angled trigonometry, and the reverse uses Pythagoras and the tangent ratio (taking care over the quadrant).

Convert a polar equation

Convert $r = 2\cos\theta$ to Cartesian form.

Step 1: Multiply through by $r$ to introduce usable Cartesian quantities

The equation contains $r$ and $\cos\theta$ separately, but our substitution rules only give us $r^2 = x^2 + y^2$ and $r\cos\theta = x$ . Multiplying both sides by $r$ creates exactly those combinations on the right-hand side.

r^2 = 2r\cos\theta.

Step 2: Substitute the Cartesian identities

Replace $r^2$ with $x^2 + y^2$ and $r\cos\theta$ with $x$ .

x^2 + y^2 = 2x.

Step 3: Rearrange by completing the square

Collect the $x$ terms and complete the square to reveal the standard form of a circle.

(x - 1)^2 + y^2 = 1.

Final answer: the curve is a circle of radius $1$ centred at $(1, 0)$ .

Sketching polar curves

To sketch a polar curve, build a table of $r$ against $\theta$ at key angles ( $0, \frac{\pi}{4}, \frac{\pi}{2}, \pi$ and so on), note where $r = 0$ (the curve passes through the pole) and where $r$ is maximal, and exploit symmetry. A curve symmetric in the initial line satisfies $f(-\theta) = f(\theta)$ . The cardioid $r = a(1 + \cos\theta)$ is a heart shape with a cusp at the pole when $\theta = \pi$ and maximum $r = 2a$ at $\theta = 0$ . The Archimedean spiral $r = a\theta$ winds steadily outwards, gaining $2\pi a$ in radius per revolution.

Tangents and area

For tangents to the curve in particular directions, write the Cartesian coordinates $x = r\cos\theta$ and $y = r\sin\theta$ as functions of $\theta$ and differentiate. A tangent parallel to the initial line is where $y$ is stationary, and perpendicular where $x$ is stationary.

Find a tangent parallel to the initial line

For the cardioid $r = 1 + \cos\theta$ , find where the tangent is parallel to the initial line.

Step 1: Express the $y$ -coordinate as a function of $\theta$

A tangent is parallel to the initial line (the $x$ -axis) when $y = r\sin\theta$ is stationary with respect to $\theta$ . We first expand $y$ using the product rule and a double-angle identity to get a form we can differentiate cleanly.

y = r\sin\theta = (1 + \cos\theta)\sin\theta = \sin\theta + \sin\theta\cos\theta = \sin\theta + \tfrac{1}{2}\sin 2\theta.

Step 2: Differentiate and factor

Differentiate $y$ with respect to $\theta$ and simplify using the double-angle identity $\cos 2\theta = 2\cos^2\theta - 1$ , which turns the expression into a quadratic in $\cos\theta$ that factors neatly.

\frac{dy}{d\theta} = \cos\theta + \cos 2\theta = \cos\theta + (2\cos^2\theta - 1) = 2\cos^2\theta + \cos\theta - 1 = (2\cos\theta - 1)(\cos\theta + 1).

Step 3: Solve $\frac{dy}{d\theta} = 0$ and identify the relevant solutions

Setting the derivative to zero gives two cases.

$\cos\theta = \tfrac{1}{2}$ : this gives $\theta = \tfrac{\pi}{3}$ or $\theta = \tfrac{5\pi}{3}$ , which are the genuine horizontal tangents. $\cos\theta = -1$ : this gives $\theta = \pi$ , which is the cusp at the pole where $r = 0$ , not a smooth tangent.

Final answer: the tangents parallel to the initial line occur at $\theta = \dfrac{\pi}{3}$ and $\theta = \dfrac{5\pi}{3}$ .

Area swept by a polar curve

Find the area enclosed by one loop where $r = 2\sin\theta$ , swept as $\theta$ runs from $0$ to $\pi$ .

Step 1: Set up the polar area integral

The area enclosed by a polar curve between two angles is given by $A = \frac{1}{2}\int r^2\,d\theta$ . Substituting $r = 2\sin\theta$ with limits $0$ to $\pi$ gives the integral for one complete loop.

A = \frac{1}{2}\int_0^{\pi} (2\sin\theta)^2\,d\theta = \frac{1}{2}\int_0^{\pi} 4\sin^2\theta\,d\theta = 2\int_0^{\pi}\sin^2\theta\,d\theta.

Step 2: Linearise $\sin^2\theta$ using a double-angle identity

We cannot integrate $\sin^2\theta$ directly, so we use the identity $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$ to replace the squared term with a sum of simpler functions.

2\int_0^{\pi}\sin^2\theta\,d\theta = 2\int_0^{\pi}\frac{1 - \cos 2\theta}{2}\,d\theta = \int_0^{\pi}(1 - \cos 2\theta)\,d\theta.

Step 3: Evaluate the definite integral

Integrate term by term and substitute the limits. The $\sin 2\theta$ term vanishes at both $\theta = 0$ and $\theta = \pi$ , leaving only the contribution from the linear term.

\left[\theta - \frac{1}{2}\sin 2\theta\right]_0^{\pi} = \left(\pi - 0\right) - \left(0 - 0\right) = \pi.

Final answer: the area enclosed is $\pi$ . This makes sense: the curve $r = 2\sin\theta$ is in fact a circle of radius $1$ centred at $(0, 1)$ , which has area $\pi$ .

Examples in context

Polar coordinates connect to several Further Maths threads. The modulus-argument form of a complex number $z = r(\cos\theta + i\sin\theta)$ is literally a polar coordinate, so de Moivre's theorem and polar curves share the same $(r, \theta)$ language. The polar area formula is the curved-region analogue of the volume and area integrals in further calculus, and integrating $r^2$ almost always requires the double-angle identities $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$ and $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$ . Conics expressed in polar form (with the focus at the pole) appear in the further coordinate systems option and in orbital mechanics.

Try this

Q1. Find the area enclosed by $r = 2$ for $0 \le \theta \le 2\pi$ . [2 marks]

Cue. $A = \frac{1}{2}\int_0^{2\pi} 4\,d\theta = 4\pi$ , the area of a circle of radius $2$ .

Q2. Convert the point with polar coordinates $\left(2, \frac{\pi}{3}\right)$ to Cartesian form. [2 marks]

Cue. $x = 2\cos\frac{\pi}{3} = 1$ , $y = 2\sin\frac{\pi}{3} = \sqrt{3}$ .

Q3. Convert $r = 3\cos\theta$ to Cartesian form. [3 marks]

Cue. $r^2 = 3r\cos\theta$ gives $x^2 + y^2 = 3x$ , i.e. $\left(x - \frac{3}{2}\right)^2 + y^2 = \frac{9}{4}$ .

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20196 marksThe curve

C

has polar equation

r = a(1 + \cos\theta)

for

0 \le \theta < 2\pi

, where

a > 0

. Find the exact area enclosed by

C

Show worked answer →

Apply the polar area formula and use the double-angle identity to integrate $\cos^2\theta$ .

$A = \frac{1}{2}\int_0^{2\pi} r^2\,d\theta = \frac{1}{2}\int_0^{2\pi} a^2(1 + \cos\theta)^2\,d\theta$ (M1).

Expand: $(1 + \cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta = 1 + 2\cos\theta + \frac{1 + \cos 2\theta}{2}$ (M1 A1).

So $A = \frac{a^2}{2}\int_0^{2\pi}\left(\frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta\right)d\theta$ . Over $0$ to $2\pi$ the $\cos\theta$ and $\cos 2\theta$ terms integrate to zero (M1 A1), leaving $A = \frac{a^2}{2}\cdot\frac{3}{2}\cdot 2\pi = \frac{3\pi a^2}{2}$ (A1).

Edexcel 20215 marksThe curve has polar equation

r = 1 + 2\cos\theta

. Find the polar coordinates of the points where the tangent to the curve is perpendicular to the initial line.

Show worked answer →

Tangents perpendicular to the initial line occur where $\frac{d}{d\theta}(r\cos\theta) = 0$ .

$r\cos\theta = (1 + 2\cos\theta)\cos\theta = \cos\theta + 2\cos^2\theta$ (M1).

Differentiate: $\frac{d}{d\theta}(r\cos\theta) = -\sin\theta - 4\cos\theta\sin\theta = -\sin\theta(1 + 4\cos\theta)$ (M1 A1).

Set to zero: $\sin\theta = 0$ or $\cos\theta = -\frac{1}{4}$ (A1). Taking $\cos\theta = -\frac{1}{4}$ gives $\theta = \arccos(-\frac14) \approx 1.823$ and its reflection, with $r = 1 + 2(-\frac14) = \frac12$ (A1).

Edexcel 20233 marksConvert the polar equation

r = 4\sin\theta

to Cartesian form and describe the curve.

Show worked answer →

Multiply by $r$ to introduce $r^2$ and $r\sin\theta$ , then substitute.

$r = 4\sin\theta \Rightarrow r^2 = 4r\sin\theta$ (M1). Using $r^2 = x^2 + y^2$ and $r\sin\theta = y$ : $x^2 + y^2 = 4y$ (A1).

Completing the square: $x^2 + (y - 2)^2 = 4$ , a circle of radius $2$ centred at $(0, 2)$ (A1).

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Further Mathematics (9FM0) specification — Pearson Edexcel (2017)